Consider the differential equation frac{dy}{d | vedclass

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(B) Given differential equation is $\frac{dy}{dx} = \frac{y^3}{2(xy^2 - x^2)}$.Let $z = y^2$. Then $\frac{dz}{dx} = 2y \frac{dy}{dx}$.Substituting $\f

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Statement $-1$ is true and statement $-2$ is false.

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(B) Given differential equation is $\frac{dy}{dx} = \frac{y^3}{2(xy^2 - x^2)}$.Let $z = y^2$. Then $\frac{dz}{dx} = 2y \frac{dy}{dx}$.Substituting $\frac{dy}{dx} = \frac{1}{2y} \frac{dz}{dx}$ into the equation:$\frac{1}{2y} \frac{dz}{dx} = \frac{y^3}{2(xy^2 - x^2)} \implies \frac{dz}{dx} = \frac{y^4}{xy^2 - x^2} = \frac{z^2}{xz - x^2}$.Dividing numerator and denominator by $x^2$,we get $\frac{dz}{dx} = \frac{(z/x)^2}{(z/x) - 1}$. This is a homogeneous differential equation in terms of $z$ and $x$. Thus,Statement $-1$ is true.To solve $\frac{dz}{dx} = \frac{z^2}{xz - x^2}$,let $z = vx$,then $\frac{dz}{dx} = v + x \frac{dv}{dx}$.$v + x \frac{dv}{dx} = \frac{v^2 x^2}{x(vx) - x^2} = \frac{v^2}{v - 1}$.$x \frac{dv}{dx} = \frac{v^2}{v - 1} - v = \frac{v^2 - v^2 + v}{v - 1} = \frac{v}{v - 1}$.$\int \frac{v - 1}{v} dv = \int \frac{1}{x} dx \implies \int (1 - \frac{1}{v}) dv = \ln|x| + C$.$v - \ln|v| = \ln|x| + C \implies \frac{z}{x} - \ln|\frac{z}{x}| = \ln|x| + C$.$\frac{y^2}{x} - \ln|y^2| + \ln|x| = \ln|x| + C \implies \frac{y^2}{x} = \ln|y^2| + C$.This does not match $y^2 e^{-y^2/x} = C$. Thus,Statement $-2$ is false.

Consider the differential equation $\frac{dy}{dx} = \frac{y^3}{2(xy^2 - x^2)}$.
Statement $-1:$ The substitution $z = y^2$ transforms the above equation into a first-order homogeneous differential equation.
Statement $-2:$ The solution of this differential equation is $y^2 e^{-y^2/x} = C$.

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Consider the differential equation $\frac{dy}{dx} = \frac{y^3}{2(xy^2 - x^2)}$.Statement $-1:$ The substitution $z = y^2$ transforms the above equation into a first-order homogeneous differential equation.Statement $-2:$ The solution of this differential equation is $y^2 e^{-y^2/x} = C$.